# Thread: Join Complete Sufficient Statistic

1. ## Join Complete Sufficient Statistic

Let X1,...,Xn be from a $N(\mu,4 \sigma^2)$

A) Find the Joint complete sufficient statistics for $\mu$ and $4 \sigma^2$

B)Find an MVUE for $4 \sigma^2$

C)Find an MVUE for $\mu^2$

a)From a previous problem we did in class, I know $Y_1 = \Sigma X^2_i$ and $Y_2 = \Sigma X_i$ are Joint complete sufficient statistics for $\mu$ and $4 \sigma^2$

b)Let $Z_2 = \frac{Y_1 - Y^2_2/n}{n-1} = \frac{\Sigma(X_i - \overline{X})^2}{n-1}$ which is the sample variance.

Would $E(Z_2) = 4 \sigma^2$ or would it be $E(Z_2) = \sigma^2$

c)Let $Z_1 = \frac{Y_2}{n} = \overline{X}$ and $E(Z_1)=\mu$

$E(Z^2_1) = E(Z_1)^2 + Var(Z_1) = \mu^2 +\frac{4 \sigma^2}{n}$

Therefore $Z^2_1 - \frac{Z_2}{n}$ is an MVUE for $\mu^2$ according to Lehmann and Scheffe Theroem.

Did I make a mistake somewhere, and would $E(Z_2) = 4 \sigma^2$ or would it be $E(Z_2) = \sigma^2$

2. You left the ^ off your Z2

Originally Posted by statmajor
Let X1,...,Xn be from a $N(\mu,4 \sigma^2)$

A) Find the Joint complete sufficient statistics for $\mu$ and $4 \sigma^2$

B)Find an MVUE for $4 \sigma^2$

C)Find an MVUE for $\mu^2$

a)From a previous problem we did in class, I know $Y_1 = \Sigma X^2_i$ and $Y_2 = \Sigma X_i$ are Joint complete sufficient statistics for $\mu$ and $4 \sigma^2$

b)Let $Z_2 = \frac{Y_1 - Y^2_2/n}{n-1} = \frac{\Sigma(X_i - \overline{X})2}{n-1}$ which is the sample variance.

Would $E(Z_2) = 4 \sigma^2$ or would it be $E(Z_2) = \sigma^2$

c)Let $Z_1 = \frac{Y_2}{n} = \overline{X}$ and $E(Z_1)=\mu$

$E(Z^2_1) = E(Z_1)^2 + Var(Z_1) = \mu^2 +\frac{4 \sigma^2}{n}$

Therefore $Z^2_1 - \frac{Z_2}{n}$ is an MVUE for $\mu^2$ according to Lehmann and Scheffe Theroem.

Did I make a mistake somewhere, and would $E(Z_2) = 4 \sigma^2$ or would it be $E(Z_2) = \sigma^2$

3. So would would $E(Z_2) = 4 \sigma^2$?

4. yes, $E(Z_2) = V(X_1)=4 \sigma^2$

5. Thank you.